I have to prove the following fact.

Show that $X_1$ is integrable, iff for all $\epsilon>0$ $$\sum_{n=1}^{\infty} \mathbb{P}(|X_1|>n \epsilon)<\infty.$$

Here $X_1$ is just a random variable (not necessarily having discrete values). And I know the following two facts.

$$\mathbb{E}X = \int_0^{\infty} \mathbb{P}(X \geq x) \, dx \qquad \quad \mathbb{E}X = \sum_{i=1}^{\infty} \mathbb{P}(X \geq i)$$

The left side is for continuous nonnegative random variable, and the right side is for discrete nonnegative random variable.

What is so frustrating here is that I have to prove the above equivalent relation regardless of $X_1$ being continuous or discrete...I can't think of an idea to start....Could anyone please help me with this?

  • $\begingroup$ This looks like a possible Borel-Cantelli application $\endgroup$ – user237393 May 5 '15 at 8:15
  • 1
    $\begingroup$ Could you give me some helps? I tried a number of times and now my brain just got stuck. $\endgroup$ – Keith May 5 '15 at 8:35
  • $\begingroup$ Please have a look at the MathJax-Tutorial $\endgroup$ – saz May 5 '15 at 14:40

First of all, note that the identity

$$\mathbb{E}X = \int_0^{\infty} \mathbb{P}(X \geq x) \, dx \tag{1}$$

holds for any non-negative random variable $X$.

Hints: Let $X$ be a non-negative random variable and $\epsilon>0$.

  1. Since $x \mapsto \mathbb{P}(X \geq x)$ is decreasing, we have $$\begin{align*} \frac{1}{\epsilon} \mathbb{E}X = \mathbb{E}\frac{X}{\epsilon} &\stackrel{(1)}{=} \sum_{n \in \mathbb{N}_0} \int_n^{n+1} \mathbb{P}(X/\epsilon \geq x) \, dx \\ &\geq \sum_{n \in \mathbb{N}_0} \mathbb{P}(X \geq \epsilon (n+1)). \end{align*}$$
  2. Conclude from the first step that $X_1 \in L^1$ implies $\sum_n \mathbb{P}(|X_1|>\epsilon n) < \infty$ for all $\epsilon>0$.
  3. Using again monotonocity, we find $$\begin{align*} \mathbb{E}X &= \sum_{n=0}^{\infty} \int_n^{n+1} \mathbb{P}(X>r) \, dr \leq \sum_{n \in \mathbb{N}_0} \mathbb{P}(X>n).\end{align*}$$
  4. Conclude that $\sum_{n \geq 1} \mathbb{P}(|X_1|>n) <\infty$ implies $X_1 \in L^1$.
  • $\begingroup$ In 3. how does $\sum_{n≥1} P(|X|>n) ≤ E[|X|]$ tell us that that $E[|X|]$ is in $L^1$? Shouldn't we be trying to find a ≥ condition? $\endgroup$ – user237393 May 5 '15 at 10:06
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    $\begingroup$ @user237393 Ah, sorry, the estimate is the wrong way round. I'll correct it. $\endgroup$ – saz May 5 '15 at 10:31

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